Question

A company manufactures two types of toys A and B. Type A requires 5 minutes each for cutting and 10 minutes each for assembling. Type B requires 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours available for cutting and 4 hours available for assembling in a day. The profit is Rs 50 each on type A and Rs 60 each on type B. How many toys of each type should the company manufacture in a day to maximize the profit?

Answer 1

Let x toys of type A and y toys of type B were manufactured.

The given information can be tabulated as follows:

	Cutting time (minutes)	Assembling time (minutes)
Toy A(x)	5	10
Toy B(y)	8	8
Availability	180	240

The constraints are
$$\begin{gathered} 5x+8y\le 180 \\ 10x+8y\le 240 \end{gathered}$$

The profit is Rs 50 each on type A and Rs 60 each on type B. Therefore, profit gained on x toys of type A and y toys of type B is Rs 50x and Rs 60 y respectively.
Total profit = Z = $50x+60y$

The mathematical formulation of the given problem is

Max Z = $50x+60y$

subject to

$$\begin{gathered} 5x+8y\le 180 \\ 10x+8y\le 240 \end{gathered}$$

First we will convert inequations into equations as follows:
5x + 8y = 180, 10x + 8y = 240, x = 0 and y = 0

Region represented by 5x + 8y ≤ 180:
The line 5x + 8y = 180 meets the coordinate axes at A₁(36, 0) and $B_1\left(0,\frac{45}{2}\right)$ respectively. By joining these points we obtain the line 5x + 8y = 180. Clearly, (0,0) satisfies the 5x + 8y = 180. So, the region which contains the origin represents the solution set of the inequation 5x + 8y ≤ 180.

Region represented by 10x + 8y ≤ 240:
The line 10x + 8y = 240 meets the coordinate axes at C₁(24, 0) and $D_1\left(0,30\right)$ respectively. By joining these points we obtain the line 10x + 8y = 240. Clearly (0,0) satisfies the inequation 10x + 8y ≤ 240. So,the region which contains the origin represents the solution set of the inequation 10x + 8y ≤ 240.

Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.
The feasible region determined by the system of constraints 5x + 8y ≤ 180, 10x + 8y ≤ 240, x ≥ 0 and y ≥ 0 are as follows.

The feasible region is shown in the figure


The corner points are B₁$\left(0,\frac{45}{2}\right)$, E₁(12, 15) and C₁(24, 0).

The values of Z at the corner points are

Corner points	Z = $50x+60y$
O	0
B1	1350
E1	1500
C1	1200

The maximum value of Z is 1500 which is at E₁(12, 15).

Thus, for maximum profit, 12 units of toy A and 15 units of toy B should be manufactured.